Last two digits of $9^{1500}$

Question

I've read this PDF where it explains how to find the last digit of a number.

If I were to find the last digit of $9^{1500}$ I would simply write it as $(3^{2})^{1500}$ and then use the patterns in the PDF for $3^{4n}$.

The problem here is that I'm asked to find the last $2$ digits. I think I could try to find patterns for the last 2 digits of $3^x$ or for $9^x$ but this would waste a lot of time, and since this problem was supposed to be solved by hand, I think this is not the best method. I'm also having problems to find literature about these problems of finding last digits of large exponents. Can somebody recommend be some?

The $9$ is nice, imagine expanding $(10-1)^{1500}$ using the binomial theorem. — André Nicolas, May 29 '15 at 06:10

score 13 · Answer 1 · answered May 29 '15 at 06:10

13

$(10-1)^{1500} = 10^{1500} - \binom{1500}{1}10^{1499}{1^1} + ... + \binom{1500}{1498}10^21^{1498} - \binom{1500}{1499}10^11^{1499} + 1^{1500}$ Only the last term is not divisible by hundred. Thus, $9^{1500} \equiv 1 \mod 100$ $i.e.$ the last two digits are 01.

answered May 29 '15 at 06:10

Ojas

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To be clearer, \begin{align}(10-1)^{1500} &= 10^{1500} - \binom{1500}{1}10^{1499}{1^1} + ... + \binom{1500}{1498}10^21^{1498} - \binom{1500}{1499}10^11^{1499} + 1^{1500}\&=100R-1500\cdot10^1\cdot1^{1499}+1\&=100(R-150)+1\&=100S+1,S\in{\Bbb{N}} \end{align} – Mythomorphic May 29 '15 at 06:20

5xum · Accepted Answer · 2015-05-29T06:17:16.823

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Calculating the last two digits is like taking the number $\mod 100$.

$$9 \equiv 9\mod 100\\ 9^2 \equiv 81\mod 100\\ 9^3=729\equiv 29\mod 100\\ 9^4=9\cdot 9^3 \equiv 9\cdot 29 = 261\equiv 61\mod 100\\ 9^5\equiv 9\cdot 61 = 549\equiv 49\mod 100\\ 9^6 \equiv 9\cdot 49 = 441\equiv 41\mod 100\\ 9^7\equiv 9\cdot 41 = 369\equiv 69\mod 100\\ 9^8\equiv 9\cdot 69 = 721\equiv 21\mod 100\\ 9^9\equiv 9\cdot 21 = 189\equiv 89\mod 100\\ 9^{10}\equiv 9\cdot 89 = 801\equiv 1\mod 100\\ $$

Now, knowing that $9^{10}\equiv 1\mod 100$, your remainging task should be easier.

edited May 29 '15 at 06:17

answered May 29 '15 at 06:13

5xum

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You made a typo in the last step. $9^{10} \equiv 9 \cdot 89.$ – Ojas May 29 '15 at 06:16
@Ojas Thanks, fixed it. – 5xum May 29 '15 at 06:17
But raising both sides by $150$ will get me $1$. I should get 2 digits. What's happening? – Poperton May 29 '15 at 06:20
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@GuerlandoOCs Well, look at it this way: what are the last two digits of $3891201$? – 5xum May 29 '15 at 06:24
@GuerlandoOCs No need to use brute-force when there are simple power(ful) lifting laws - see my answer. – Bill Dubuque Apr 11 '19 at 19:41

score 2 · Answer 3 · answered May 29 '15 at 06:10

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HINT:

$$9^{2n}=(10-1)^{2n}=(1-10)^{2n}\equiv1-\binom{2n}110^1\pmod{100}\equiv1-20n$$

$\implies9^{2n}\equiv1\pmod{100}$ if $5|n$

Else $1-20n\equiv1+80n\pmod{100}$

answered May 29 '15 at 06:10

lab bhattacharjee

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wlad · Answer 4 · 2019-04-11T06:47:32.033

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$$\phi(100) = 40 \therefore 9^{1500} \equiv (9^{40})^{37}9^{20} \equiv 9^{20} \equiv \pm1, \pm51 \pmod{100}$$

as these all square to $1$ mod 100. To find which of the four is the right one: $$9^{1500} \equiv 81^{750} \equiv 1^{750}\equiv 1 \pmod{20} $$ So $01$.

edited Apr 11 '19 at 06:47

answered May 29 '15 at 06:12

wlad

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$$\phi(100)=\phi(4)\cdot\phi(25)=2\cdot20,$$ right? http://en.wikipedia.org/wiki/Carmichael_function will be more useful here – lab bhattacharjee May 29 '15 at 06:14
@labbhattacharjee I think I've addressed it. – wlad May 29 '15 at 06:25
Clever enough >Never thought this way – lab bhattacharjee May 29 '15 at 06:34
(+1) I didn't see your answer before I posted, but I have a bit more explanation, so I will leave mine unless you object. – robjohn May 29 '15 at 07:37
How did you deduce $!\bmod 100!:\ x\equiv 9^{20}\equiv \pm1,$? $ $ Note $,x^2-1,$ has $2$ more roots $,x\equiv \pm 51\ $ – Bill Dubuque Apr 10 '19 at 22:51
@BillDubuque Good point. I've included the extra roots and a calculation mod 20. I could add that the roots could be computed using CRT – wlad Apr 11 '19 at 06:50
That fixes it but, alas, now it's much longer since you have to use (unshown) CRT etc (vs. e.g. short way in my answer) – Bill Dubuque Apr 11 '19 at 19:18

score 2 · Answer 5 · answered May 29 '15 at 07:12

$$9^2\equiv -19\pmod{100}$$

$$\implies 9^4\equiv 361\pmod{100}$$

$$\implies9^4\equiv 61\pmod{100}$$

$$\implies9^8\equiv 3721\pmod{100}$$

$$\implies 9^8\equiv21\pmod{100}$$

$$\implies9^8.9^{2}\equiv(21.81)\pmod{100}$$

$$\implies9^{10}\equiv1701\pmod{100}$$

$$\implies9^{10}\equiv 01\pmod{100}$$

$$\implies9^{1500}\equiv 01\pmod{100}$$

score 2 · Answer 6 · answered May 29 '15 at 07:35

Since $\phi(100)=\phi(2^2\cdot5^2)=1\cdot2^1\cdot4\cdot5^1=40$ (where $\phi$ is Euler's Totient), we know that $$ x^{40}\equiv1\pmod{100} $$ Therefore, since $1500\equiv20\pmod{40}$ $$ 9^{1500}\equiv9^{20}\pmod{100} $$ and since $20=10100_\text{two}$, we can compute by squaring and multiplying $$ \begin{align} 9^2&\equiv81\\ 9^4&\equiv61\\ 9^5&\equiv49\\ 9^{10}&\equiv1\\ 9^{20}&\equiv1\pmod{100} \end{align} $$ Thus, the last two digits of $9^{1500}$ are '$01$'.

Bill Dubuque · Answer 7 · 2021-05-22T22:40:03.800

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Hint $\ 9\equiv -1\pmod{\!10}\,\Rightarrow\,9^{\large 10}\equiv (-1)^{\large 10}\,\pmod{\!10^{\large 2}}\ $

since $\,\ \ a\,\equiv\ b\, \pmod{\!n}\ \Rightarrow\,\ a^{\large n}\,\equiv\ b^{\large n}\ \ \pmod{\!n^{\large 2}},\ $ [simple proof here]

edited May 22 '21 at 22:40

answered Apr 11 '19 at 19:14

Bill Dubuque

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J.G.131 · Answer 8 · 2024-05-20T21:51:41.023

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Since $\phi(100)\ |\ 3000$, it follows that $9^{1500}=$$(3^2)^{1500}=3^{3000}\equiv 1(\text{mod}(100))$.

edited May 20 '24 at 21:51

answered May 16 '24 at 15:35

J.G.131

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Special case of the Remark in the dupe. – Bill Dubuque May 20 '24 at 22:05

Last two digits of $9^{1500}$

8 Answers8